Geothermal VRF replaces gas boilers for pool heating

International Research Journal of Engineering and Technology (IRJET) e-ISSN: 2395-0056
Volume: 09 Issue: 03 | Mar 2022 www.irjet.net p-ISSN: 2395-0072
© 2022, IRJET | Impact Factor value: 7.529 | ISO 9001:2008 Certified Journal | Page 255
REPLACEMENT OF GAS BOILEIR BY GEOTHERMAL/AEROTHERMAL
VRF-HEATPUMP FOR SWIMMING POOL HEATING IN MOROCCO.
Karime MAROUF1, Said HACHIMI2
1Karime Marouf, Applied Thermodynamics and Solid Fuels, Mohammedia School of Engineering, University
Mohammed V, Rabat Morocco.
2Said Hachimi, Applied Thermodynamics and Solid Fuels, Mohammedia School of Engineering, University
Mohammed V, Rabat Morocco.
---------------------------------------------------------------------***----------------------------------------------------------------------
Abstract - Faced with the challenges of the rapid increase in
energy consumption in Morocco, where more than 95% of
energy needs are imported from outside, energy efficiency has
become one of the key issues in all areas of activity. In the
latter, the energy needs are very varied and the heating
systems represent a significant part of the electricity
consumption. According to the AMEE [1], Morocco has set
itself the goal of achieving energy savings of 15% by 2030,
through the implementation of an energyefficiencyplan in the
various economic sectors. This is particularly the heating
systems.
To respond to this situation, we replaced a gas boiler system
with ground source VRF (variablerefrigerantflow)heatpump
mixed with aerothermal VRF for heating an outdoor
swimming pool in Rabat (Morocco). The objective of this
article is to analyze the electrical consumption of the new
installation and define the real ROI based on two years
running of this pilot site.
Key Words: Swimming pool heating, very low enthalpy
geothermal energy, VRF, Geothermal, Aerothermal,
Electricity consumption, ROI.
1. INTRODUCTION
Morocco is an energy importing country because it has
virtually no natural resources to meet its energy needs.
Over the past decade, the country's energy needs have
increased significantly. This is mainly due to the
industrialization of the country and the good global
economic dynamics.
In view of these elements, the exploitation of geothermal
resources in Morocco allows the production of thermal
energy in a sustainable way for the heating of swimming
pools.
2. EXISTING INSTALLATION
Most of the heat used for non-electric heating results from
the combustion of hydrocarbons. The combustion of gas
mainly emits water vapor and carbon dioxide (CO2). Thus,
Morocco notes a weakness of non-renewable resources.
The system is equipped with two propane gas boilers with
an output of 800 kW.
- Annual propane consumptionisapproximately81.4tonnes
- The average price of propane in Morocco duringtheyear of
operation is 12.9 MAD / kg
- The price paid for one year is: 1 050 000 MAD (Moroccan
dirham)
3. NEW INSTALLATION
A heat pump (VRF) is a thermal machine that uses
mechanical energy (compressor pumps) to extract low-
temperature heat from an environment and restore it at a
higher temperature to make it usable for heating needs.
We have installed a mix system between 30% geothermal
heat pump and 70% aerothermal heat pump to guarantee
energy performance.
The set temperature is 29 °C
The well is at 22 m, with a temperature of 15°C and a
capacity of 20 m3/h
To estimate the calorific power to beinstalled,a heatbalance
must be carried out takingintoaccountthemostunfavorable
external conditions of the year:
Thermal balance during the filling phase in order to deduce
the calorific power required during filling P1.
Thermal balance enabling the calorific power required to
maintain the temperature of the pool water at 28°C during
the operating period P2 to be deduced.
With:
Estimation of the heating capacity during the filling
phase:
P1 = Surface losses + Wall heat losses + Water and concrete
heating - Solar gains
-Surface losses = Losses by evaporation + losses by
convection + losses by radiation

Evaporation loss- source ASHRAE 2019:
When the vapor pressure of the ambientairislowerthanthe
saturated vapor pressure at the surface of the water,
evaporation of water occurs.
In the literature, several scientific formulas deduced from
experiments make it possible to quantify losses by
evaporation.
In this project, I chose to use the formula from the American
ASHRAE 2019 [2]:
kW   
0,089 0,0782 0,5
evap water w a
P S P P V
    
kPa %
a w
P HR P
 
With:
0.5: is a coefficient linked to the activity in the basin. During
the filling phase, there is no swimmer. We then multiplybya
coefficient of 0.5, as shown in the following table.
The following activity factors should be applied to the areas
of specific features, and not to the entire wetted area:
Outside temperature when filling: 10 °C
Desired water temperature (setpoint): 28 °C
Ambient air speed at 1m from the pool surface: V = 1.2 m/s
Pw (saturated vapor pressure): at T = 28 °C, Pw = 3.783 kPa
Pa (saturation pressure): at T = 10 °C and for relative
humidity RH = 70%;
kPa 0,7 1,228 0,86
a
P   
Swater= 340.12 m2 (Water surface)
kW 90,89
evap
P 
Convection loss:
Convection is one of the thermal heat transfer modes. As the
air above the pool moves at any speed, the water molecules
also move, causing thermal heat transfer in the event that
there is a temperature gradient.
Also, several scientific experiments have been made in this
direction, in order to present a formula for quantifying the
heat exchange coefficient. Theformula usedinthis study was
also taken from the ASHRAE.
kW   3
10
conv water C water amb
P S h T T 
   
W/m2.K 5,7 3,8
c
h V
  
For a wind speed of V = 1.2 m / s
kW 62,81
conv
P 
Radiation loss:
Everything emits radiation of an electromagnetic nature,
whatever its state: solid, liquid or gas. This emission of
energy comes at the expense of the body's internal energy.
Water exchanges with the environment by radiative heat
transfer.
This radiative exchange is expressed by the following
relation:
kW
8 4 4 3
5,68 10 ( 273) ( 273) 10
ray water water water sky amb
P S T T
 
 
 
      
 
0,004 0,8
sky dew
T
   
Digital Application:
For Tamb = 10 °C and RH = 70%, T dew = 4.79 °C therefore
0,82
sky
 
kW 49,04
ray
P 
Surface losses = 202.74 kW
-Losses through the walls of the basin:
Heat losses represent heat losses through the walls of the
basin. They are all the more important when the walls are
poorly thermally insulated.
The thermal losses are expressed as follows:
kW   3
10
water earth
Q U S T 
    

The surface of the concrete walls: 143 m2
Tearth: 15 °C, temperature of the earth in winter.
Twater: 28 °C
U: the transmission coefficient of concrete is around 2.5 W /
m2.K, but for safety margin, we take a coefficient of 3.5 W /
m2.K
kW 6,5
Q 
-Water and walls heating:
To reach the set temperature in the basin, it is necessary to
raise the temperature of the water and the walls of the
concrete basin, from their initial temperaturestothedesired
temperature.
Water heating:
According to the calorimetry relationship, the capacity
required to heat the volume of water in 72 hours is as
follows:
kW
 
3600
w
w w P finalwater initialw t
wat
a er
er
P
V C T T
h
   


kW 197,38
water
P 
Heating of the walls:
According to the calorimetry relationship, the capacity
required to heat the walls of the basin made especially of
concrete for a period of 48 hours is as follows:
kW
 
3600
c
c c P finalconcrete initialconcrete
wall
P
V C T T
h
   


Concrete volume: S concrete x concrete thickness = 57.2 m3,
with concrete thickness = 0.4 m
Concrete Cp: 0.837 KJ / Kg.K
Concrete density = 2500 kg / m3
Initial concrete T (average): 9.5 ° C
T final concrete (average) = ((Hotwatertemperature+ earth
temperature) / 2)
kW
 
2500 57,2 0,837 21,5 10
72 3600
wall
P
  


kW 5,31
wall
P 
PHeating = Pwater+Pwalls
PHeating = 202.69 Kw
-Solar gains (in winter) = 0 W
P1 = 411.93kW
Estimation of the heating capacity during the
maintenance phase:
P2 = Surface losses + Wall heat losses + Renewed water
heater - Solar gains
-Surface losses = Losses by evaporation + losses by
convection + losses by radiation
Evaporation loss- source ASHRAE 2019:
As a result of the activity of bathers, evaporation losses tend
to increase. It is therefore necessary to take into account the
impact of the activity of bathers during this phase.
According to the ASHRAE 2019, for public swimming pools,
the equation must be multiplied by the activity factor Fa
(Factor Activity): 1
kW   
0,089 0,0782 1
evap water w a
P S P P V
    
We use the unfavorable climatic conditions:
Outdoor temperature: 4 °C
Desired water temperature (setpoint): 28 °C
Ambient air speed at 1m from the pool surface: V= 1.2 m/s
Pw (saturated vapor pressure at hot water temperature): at
T = 28 °C, Pw = 3.783 KPa
Pa (saturation pressure at ambient air temperature):atT=4
°C and for relative humidity RH = 88%
kW 190,73
evap
P 
Losses by convection:
The unfavorable external conditions being the same; during
the temperature maintenance phase, convection losses are
calculated as follows:
kW ,75
83
conv
P 

Radiation losses:
The unfavorable external conditions being the same; during
the temperature maintenance phase, radiation losses are
calculated as follows:
kW 58,66
ray
P 
Surface losses = 333.14 kW
-Losses through the walls of the basin: kW 6,5
Q 
-Water heating renewed:
To preserve the quality and volume of water over the long
term, water must be constantlyrecycled.Accordingtoarticle
D1332-6 of the public health code in France, an intake of 30
liters per day and per bather must be respected. This rule is
also required in Morocco.
Water renewal rate per day: 30 liters per bather.
Estimated number of people per day: 90 people.
Volume of water to be renewed: Renewal rate x Number of
people.
Calculation of the volume of water to be renewed per day:
l Volume of water renewed = Renewal rate x Number of
people
l Volume of water renewed = 30 liters x 90
l Volume of water renewed = 2700 liters
m3 Volume of water renewed = 2.7 m3
The capacity required for heating the renewed water:
kW
 
3600
w
w wr P finalwater i l
r
nitia wate
en d
r
ewe
P
V C T T
h
   


kW 2,35
renewed
P 
-Solar gains (in winter) = 0 W
P2 = 341.99 kW
The capacity needed to heat the pool in 72 hours is P1 =
411.93 kW, while the temperature maintenance capacity is
P2 = 341.99 kW. The largest capacity is used to select the
equipment to be installed. Therefore, we installed a capacity
of P = 441 kW
- Installation diagram:
The entire facility contained:
- 2 geothermal unit to maketheexchangewiththe exchanger
through softened water, and with the hydro kits throughthe
refrigerant. The capacity of each outdoor unit is 63 kW.
- 5 aeraulic units to make the heat exchange with air and
with the hydro kits through the refrigerant. The calorific
power for each group is 63 kW.
- An exchanger to exchange heat between the geothermal
groups and the wells.
- 14 hydro kits for the exchange between hot water and
refrigerant.
- A submersible pump designed to fetch hot water from a
depth where it is impossible to suck it from the surface.
- A circulation pumps to float hot water in the pipes.
- The flow switches make it possible to check thepresence of
a minimum flow in the installation
- The safety valves for pressure release in case of
overheating of the installation.
- The manometers to measure the pressure.
- The thermometers to measure the temperature.
- The automatic air vents that allow the installation to
operate in the best conditions. It eliminates air, thepresence
of which disrupts heat exchange, which causes noise and
promotes circuit corrosion
- Expansion tanks to absorb variations in the volume of
water due to temperature changes in the installation.
- Anti-vibration sleeves for absorbing the vibration of
circuits and pumps.

- Non-return valves to control the direction of circulation.
- Balancing valves to control the flow and pressure.
- Drain valves.
- Shut-off valves to isolate the circuit.
- The collectors to accumulate and distribute water.
After studying the system for two years, the diagram below
clearly defines additional guidelines to optimize the
performance of the entire system.
- 2 Heat exchangers are recommended nearby swimming
pool. During maintenance, 2nd heat exchanger canallowthe
operation.
- We recommend to design heat exchanger near swimming
pool based on 40℃ inlet temp. (Originally, inlet temp. was
38℃)
- In each zone, we recommend to add a pump. During
maintenance, the pump can allow the operation.
- We recommend to use water piping expertize in order to
maintain installation quality.
- Expansion tank needs to be designed based on total piping
length and water volume.
- It is not necessary to install excessive pressure safety valve
in each pipe. We recommend to install in main line.
Hydrokit Water Piping installation :
- For the water pipe system, we used a closed loop type.
- For the drain pipe size, we used the same diameter as the
product connected or larger.
- We installed a natural drainage so that the drained water
does not back flows.
- To replace the connected device easily, we installed the
union joint.
- We installed the service port to clean the heat exchanger at
each inlet and outlet of the water pipe.
- We installed the strainer on thehorizontal pipe.(Whendirt,
trash, rusted pieces get into the water pipe system, it can
cause problems to the product by corroding the metallic
material.)
- We installed the air vent at the top of the water pipe. If the
air vent is not installed at the top of the water pipe, there
would be a lot of bubble in the water pipe. So a plate heat
exchanger burst may happen because of the reduction of
water flow rate(CH 14 is displayed in the remote controller)
caused by a lot of bubble in the water pipe.
- We installed a thermometer and pressure gaugeattheinlet
and outlet of the water pipe.
- We installed the drain valve that can be used for draining
the water inside when replacing the part or providing
service.
- We installed the shut-off valve to block thewater byclosing
the valve when replacing the part or cleaning.
- We installed excessive pressure safety valve that meetsthe
design water pressure to prevent unit or water pipe damage
at the pressure increase inside the water pipe system.

4. ROI calculation
For the ground heat source heat pumps, the heat is taken
from a well , this heat is then transferred by the heat pump
to the hot water circuit of the heating installation.
We have 2 outdoor units, the heat output of each outdoor
unit is 63 kW. [3]
The air-water heat pump is a system that relies on the use
of calories contained in the outside air to heat the water that
is present in the heating network.
We have 5 outdoor units, the heat output of each outdoor
unit is 63 kW. [4]
- The annual average consumption during two years is
approximately 204,262 kW
- The average price of electricity in Morocco is about 1.1
MAD / kWh
- The price paid for one year is: 224 688.2 MAD
- The investment cost including equipment, installation
and maintenance is: 2 000 000 MAD
 The gain in invoice in 1 year: 1 049 775.31 MAD /year
 Payback time (ROI) is approximately 1.9 years (22.86
months)
5. CONCLUSION
Very low-temperature geothermal energyinMoroccoallows
the production of thermal energy in a sustainable way.
Indeed, the installation of geothermal heat pumps insteadof
boilers gives us a return on investment time generally less
than 2 years.
REFERENCES
[1] The Thermal Regulation of Construction in Morocco
https://www.amee.ma/sites/default/files/inline-
files/guide%20chaffage%20ventilation%20et%20cilma
tisation.pdf
[2] 2019 ASHRAE HANDBOOKHeating, Ventilating,andAir-
Conditioning Applications
[3] LG MULTI V Water product Databook
[4] LG MULTI V product Databook

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